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Can’t understand function and reference behaviour

I have this code.

JavaScript

Expected Output:

JavaScript

Original Output:

JavaScript

Inside the function $counter=0 is assigned by 0 so the &$counter should be 0.
So when i called it second time it sees $counter=0 and so that &$counter will be 0, Isn’t it?
Why it is incrementing?

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Answer

It does not call $counter=0 for the second time. You call it just once when initiating the first function. When you call $add(), you call every time the second function (that is in your return statement) which just uses the modified value of $counter that you passed by reference. If you would add echo $counter; after the $counter = 0; you will see that.

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